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Lady bird [3.3K]
4 years ago
5

Determined to test the law of gravity for himself, a student walks off a skyscraper 180 m high, stopwatch in hand, and starts hi

s free fall (zero initial velocity). Five seconds later, Superman arrives at the scene and dives off the roof to save the student.
a) Superman leaves the roof with an initial velocity that he produces by pushing himself downward from the edge of the roof with his legs of steel. He then falls with the same acceleration as any freely falling body. What must the value of the initial velocity be so that Superman catches the student just before they reach the ground?
b) On the same graph, sketch the positions of the student and of Superman as functions of time. Take Superman's initial speed to have the value calculated in part (a).
c) If the height of the skyscraper is less than some minimum value, even Superman can't reach the student before he hits the ground. What is this minimum height?
Physics
1 answer:
Sati [7]4 years ago
8 0

Answer:

a)  v₀ = - 164.62 m / s , c) y = 122.5 m

Explanation:

We can solve this exercise using the free fall kinematic relations.

We put our reference system on the floor, so the height of the skyscraper is y₀ = 180m and the floor level is y = 0 m

 

For the boy

         y = y₀ + v₀ t - ½ g t²

with free fall its initial speed is zero

        y = ½ g t2

For superman

        y = y₀ + v₀ (t-5) - ½ g (t-5)²

how superman grabs the lot just before hitting the ground

we look for the time it takes the boy down

         t = √ (2 y₀ / g)

         t = √ (2 180 / 9,8)

         t = 6.06 s

in the equation for superman, we clear the volume and calculate

         v₀ (t-5) = -y₀ + ½ g (t-5)²

         v₀ (6.06 -5) = -180 + ½ 9.8 (6.06 -5)²

         v₀ 1.06 = -174.49

         v₀ = - 174.49 / 1.06

         v₀ = - 164.62 m / s

the negative sign indicates that the initial speed is down

b) to graph the position of the two we use the table

  t (s)      Y_boy (m)   Y_superman (m)

    0             180                 180

   1              175.1               180

   5              57.5              180

   6                3.6                10.18

see attachment for the two curves

c) calculate the height that falls a lot in the 5 seconds (t = 5)

           y = -1/2 g t²

           y = ½ 9.8 5²

           y = 122.5 m

for this height superman has not yet left the skyscraper, so the boy hits the ground

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8_murik_8 [283]

Answer:

Difference in Twin's Ages = 12.68 years

Explanation:

Using special theory of relativity's time dilation phenomenon, we first find the time that is passed on earth during Lou's trip.

t = t₀/[√(1 - v²/c²)]

where,

t = time measured by the person in relative motion = 1 year

t₀ = time measured by the person at rest = ?

v = speed of relative motion = 0.96 c

c = speed of light  

Therefore,

1 year = t₀/[√(1 - 0.96² c²/c²)]

1 year = t₀/[√(1 - 0.9216)]

(1 year)(0.28 year) = t₀

t₀ = 0.28 year

Let,

y = Sue's age

x = Lou's age

so,

x - y = 13.4 years

but, after this trip Lou has aged 1 year, and on earth only 0.28 years passed so, Sue has aged only 0.28 years. Therefore,

x = x + 1

y = y + 0.28

Therefore,

(x + 1 year) - (y + 0.28 year) = 13.4 years

x - y = 13.4 years - 0.72 year

x - y = 12.68 years

<u>Difference in Twin's Ages = 12.68 years</u>

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4 years ago
A 50g ball is released from rest 1.0 above the bottom of thetrack
ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

y=\dfrac{1}{4}x^2

We need to calculate the velocity

Using conservation of energy

\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

\Delta U_{i}=\Delta K_{f}

Put the value into the formula

mgh=\dfrac{1}{2}mv^2

Put the value into the formula

50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2

v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}

v=\sqrt{19.6}

v=4.42\ m/s

We need to calculate the maximum height of the ball

Using again conservation of energy

\dfrac{1}{2}mv^2=mgh

Here, h = y highest point

Put the value into the formula

\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h

y=\dfrac{0.5\times(4.42)^2}{9.8}

y=0.996\ m

Put the value of y in the given equation

y=\dfrac{1}{4}x^2

x^2=4\times0.996

x=\sqrt{4\times0.996}

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Hence, The maximum height of the ball is 2 m.

4 0
3 years ago
If an object has zero acceleration, does that mean it has zero velocity?
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A .183 kg ball is moving 18.8 m/s when it runs into a spring of spring constant 86.9 N/m. How much KE does the ball have when it
Lemur [1.5K]

Answer:

The value is  KE_B  = 20.59 \  J

Explanation:

From the question we are told that

   The mass of the ball is  m  =  183 \  kg

   The initial  speed of the ball is  u  =  18.8 \  m/s

    The spring constant is  k  =  86.9 \ N/m

     The compression distance is  x =  0.520 \ m

Generally the energy stored in the string is mathematically represented as

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=>     E =  \frac{1}{2}  *  86.9  * 0.520 ^2

=>      E =   11.75 \  J

Generally the kinetic energy of the ball is mathematically represented as

          KE_b  =  \frac{1}{2} * m * u^2

=>      KE_b  =  \frac{1}{2}  *  0.183 * (18.8 )^2

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Generally the KE   the ball have when it has compressed the spring is mathematically represented as

          KE_B  =  KE_b -  E

=>        KE_B  =  32.34 - 11.75

=>        KE_B  = 20.59 \  J

7 0
3 years ago
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